How do you rationalize $\frac{x}{\sqrt{5} - \sqrt{2}}$ $x/(sqrt5-sqrt2)$ ?

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How do you rationalize $\frac{x}{\sqrt{5} - \sqrt{2}}$ $x/(sqrt5-sqrt2)$ ?

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Answer 1 · 2015-05-21T22:12:42

You can rationalize the denominator of this expression, by multiplying the numerator (top) and denominator (bottom) by the conjugate of $\sqrt{5} - \sqrt{2}$ $sqrt(5)-sqrt(2)$ , namely $\sqrt{5} + \sqrt{2}$ $sqrt(5)+sqrt(2)$ .

$\frac{x}{\sqrt{5} - \sqrt{2}}$ $x/(sqrt(5)-sqrt(2))$

$= \frac{(\sqrt{5} + \sqrt{2}) x}{(\sqrt{5} + \sqrt{2}) (\sqrt{5} - \sqrt{2})}$ $= ((sqrt(5)+sqrt(2))x)/((sqrt(5)+sqrt(2))(sqrt(5)-sqrt(2)))$

$= \frac{(\sqrt{5} + \sqrt{2}) x}{\sqrt{5} \sqrt{5} - \sqrt{2} \sqrt{2}}$ $= ((sqrt(5)+sqrt(2))x)/(sqrt(5)sqrt(5)-sqrt(2)sqrt(2))$

$= \frac{(\sqrt{5} + \sqrt{2}) x}{5 - 2}$ $= ((sqrt(5)+sqrt(2))x)/(5-2)$

$= \frac{(\sqrt{5} + \sqrt{2}) x}{3}$ $= ((sqrt(5)+sqrt(2))x)/3$

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How do you rationalize $\frac{x}{\sqrt{5} - \sqrt{2}}$ $x/(sqrt5-sqrt2)$ ?

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